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bezimeni [28]
3 years ago
12

You can increase the capacitance of a capacitor by A. Decreasing the plate spacing B. Increasing the plate spacing. ° C. Decreas

ing the area of the plates. D. Increasing the area of the plates. E. Both A and D F. Both B and C
Physics
1 answer:
eimsori [14]3 years ago
4 0

You can increase the capacitance of a capacitor by decreasing the plate spacing (A) or by increasing the area of the plates (D).

'A' and 'D' both do the job, so the correct choice is<em> (E)</em> .

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The current in a wire varies with time according to the relationship 1=55A−(0.65A/s2)t2. (a) How many coulombs of charge pass a
mario62 [17]

Answer:

(A) Q = 321.1C (B) I = 42.8A

Explanation:

(a)Given I = 55A−(0.65A/s2)t²

I = dQ/dt

dQ = I×dt

To get an expression for Q we integrate with respect to t.

So Q = ∫I×dt =∫[55−(0.65)t²]dt

Q = [55t – 0.65/3×t³]

Q between t=0 and t= 7.5s

Q = [55×(7.5 – 0) – 0.65/3(7.5³– 0³)]

Q = 321.1C

(b) For a constant current I in the same time interval

I = Q/t = 321.1/7.5 = 42.8A.

3 0
3 years ago
Read 2 more answers
What is heat?
OLEGan [10]
Well you need to have lots of heat
7 0
3 years ago
A boat leaves the dock at t = 0.00 s and, starting from rest, maintains a constant acceleration of (0.461 m/s2)i relative to the
liberstina [14]

Answer:

At t=4.82 s, the boat is moving at 3.464 m/s.

At t=4.82 s, the boat is 13.112 m from the dock.

Explanation:

The speed of the boat in j'th direction remains constant for all times (vj=2.16 m/s), however, the speed in i'th direction is changing due to the constant acceleration (0.461 m/s^2)i.

In order to find the velocity of the boat a t=4.82 s, first we need to compute the velocity of the boat relative to the water in the i direction (vi_b) at t=4.82 s:

vi_b = a*t = (0.461 m/s^2)*(4.82 s) = 2.222 m/s

Now, we add this velocity to the velocity of the water in the i direction:

vi = vi_b + vi_w = 2.222 m/s + 0.486 m/s = 2.708 m/s

Therefore, the speed of the boat at t = 4.82 s is: v = (vi, vj) = (2.708, 2.16) m/s. Finally, to find its speed, we just calculate the magnitude of v and obtain that the speed is: 3.464 m/s.

For the second question, first we will find the distance that the boat moved in the i'th direction and then in the j'th direction.

The speed in the i'th direction, for all times, is given by:

(0.485 + 0.461*t) and in order to find the distance advanced in the i'th direction (di) during 4.82 s, we need to integrate this velocity:

di = 0.485*t + (0.461*t^2)/2 (evaluated from t=0 to t =4.82) = 0.485*(4.82) + (0.461*(4.82)^2)/2 = 2.337 + 5.634 = 7.971 m

The speed in j'th direction, for all times, is given by:

2.16 and in order to find the distance advanced in the j'th direction (dj) during 4.82 s, we need to integrate this velocity:

dj = 2.16*t (evaluated from t=0 to t =4.82) = (2.16)*(4.82) = 10.411 m

Using Pythagoras' Theorem, we find that the the boat is at 13.112 m from the dock at t = 4.82 s.

4 0
3 years ago
A 49.3 g ball of copper has a net charge of 2.0 µc. what fraction of the copper's electrons has been removed? (each copper atom
Serggg [28]
First, find how many copper atoms make up the ball: 
 moles of atoms = (49.3 g) / (63.5 g per mol of atoms) = 0.<span>77638</span><span>mol 
</span> # of atoms = (0.77638 mol) (6.02 × 10^23 atoms per mol) = 4.6738*10^23<span> atoms </span>

<span> There is normally one electron for every proton in copper. This means there are normally 29 electrons per atom:
</span> normal # electrons = (4.6738 × 10^23 atoms) (29 electrons per atom) = <span> <span>1.3554</span></span><span>× 10^25 electrons 
</span>
<span> Currently, the charge in the ball is 2.0 µC, which means -2.0 µC worth of electrons have been removed.
</span><span> # removed electrons = (-2.0 µC) / (1.602 × 10^-13 µC per electron) = 1.2484 × 10^13 electrons removed
 
</span><span> # removed electrons / normal # electrons = </span>
<span>(1.2484 × 10^13 electrons removed) / (1.3554 × 10^25 electrons) = 9.21 × 10^-13 </span>

<span> That's 1 / 9.21 × 10^13 </span>
7 0
3 years ago
A train is moving with a velocity of 20 m/s. when the brakes are applied the acceleration is reduced to -0.6 m/s².calculate the
Serhud [2]

Answer:

x = 333.33 [m]

Explanation:

To solve this problem we must use the following kinematics equation.

v_{f}^{2} = v_{i}^{2}-(2*a*x)

where:

Vf = final velocity = 0

Vi = initial velocity = 20 [m/s]

a = desacceleration = 0.6 [m/s^2]

x = distance [m]

Note: the final speed is zero as the body finishes its movement.

Now replacing:

0 = (20)^2 - (2*0.6*x)

1.2*x = 400

x = 333.33 [m]

3 0
3 years ago
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